Topic 9 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro’s number (NA) 1 mol = 6.02 1023 representative particles (atoms, ions, formula units, or molecules If you could spend 1 million dollars per second, how many years will it take to spend a mole of dollars, given there are 31536000 seconds per year. 1.91 x 1010 years or 19.1 billion years A. What is the Mole? 1 mole of hockey pucks would equal the mass of the moon! 1 mole of basketballs would fill a bag the size of the earth! 1 mole of pennies would cover the Earth 1/4 mile deep! A. Moles Used as a Conversion Factor: 1 mole = 6.02 x 1023 so… Moles converted to Particles = # mol 6.02 1023 Particles 1 mol A. Moles Rep. Particles converted to moles = # particles 1 mol 6.02 1023 Particles B. Molar Mass Mass of 1 mole of an element or compound. Atomic mass tells the... atomic mass units per atom (amu) grams per mole (g/mol) Round to 2 decimal places B. Molar Mass Examples carbon 12.01 g/mol aluminum 26.98 g/mol zinc 65.39 g/mol B. Molar Mass Examples water H2O 2(1.01) + 16.00 = 18.02 g/mol sodium chloride NaCl 22.99 + 35.45 = 58.44 g/mol B. Molar Mass Examples sodium hydrogen carbonate NaHCO3 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol sucrose C12H22O11 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol C. Molar Conversions molar mass 6.02 1023 MASS NUMBER MOLES IN GRAMS OF PARTICLES (g/mol) (particles/mol) C. Molar Conversion Examples How many molecules are in 2.50 moles of C12H22O11? 6.02 1023 2.50 mol molecules 1 mol = 1.51 1024 molecules C12H22O11 C. Molar Conversion Examples How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C C. Molar Conversion Examples the mass of 2.1 1024 molecules of NaHCO3. Find 2.1 1024 molecules 1 mol 84.01 g 6.02 1023 1 mol molecules = 290 g NaHCO3 Topic 9 – The Mole II. Formula Calculations I II III IV A. Percentage Composition the percentage by mass of each element in a compound mass of element % composition 100 total mass A. Percentage Composition Find %Cu = %S = the % composition of Cu2S. 127.10 g Cu 100 = 159.17 g Cu2S 79.852% Cu 32.07 g S 159.17 g Cu2S 100 = 20.15% S A. Percentage Composition Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. 28 g 100 = 78% Fe %Fe = 36 g %O = 8.0 g 36 g 100 = 22% O A. Percentage Composition How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu A. Percentage Composition Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? %H2O = 36.04 g 100 = 24.51% H2O 147.02 g B. Empirical Formula Smallest whole number ratio of atoms in a compound C 2H 6 reduce subscripts CH3 B. Empirical Formula Calculation 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s. B. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol = 1.85 mol N =1N 1.85 mol 14.01 g 74.1 g 1 mol = 4.63 mol O = 2.5 O 16.00 g 1.85 mol B. Empirical Formula N1O2.5 Need to make the subscripts whole numbers multiply by 2 N2O5 C. Molecular Formula “True Formula” - the actual number of atoms in a compound empirical formula CH3 ? molecular formula C2H6 C. Molecular Formula Calculation 1. Find the empirical formula. 2. Calculate the empirical formula molar mass. 3. Divide the molecular molar mass by the empirical molar mass. 4. Multiply each subscript by the answer from step 3. MF mass n EF mass EF n C. Molecular Formula The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.06 g/mol? empirical mass = 14.03 g/mol 28.06 g/mol 14.03 g/mol = 2.00 (CH2)2 C2H4 Topic 9 The Mole III. Molarity I II III IV A. Molarity Concentration of a solution. substance being dissolved moles of solute Molarity (M) liters of solution total combined volume A. Molarity 2M HCl What does this mean? mol M L 2 mol HCl 2M HCl 1L B. Molarity Calculations molar mass (g/mol) 6.02 1023 (particles/mol) MASS IN NUMBER MOLES OF GRAMS PARTICLES Molarity (mol/L) Aqueous solution LITERS OF SOLUTION B. Molarity Calculations How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L 0.25 mol 58.44 g 1L 0.25 mol 0.25M 1L 1 mol = 7.3 g NaCl B. Molarity Calculations Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g 1 mol mol M L M= 41.99 g 0.238 mol 0.25 L = 0.238 mol NaF = 0.95M NaF C. Molar Volume at STP 1 mol of any gas=22.4 L at STP Standard Temperature & 0°C and 1 atm Pressure C. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) 6.02 MOLES 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION NUMBER OF PARTICLES C. Molar Conversion Examples How many Liters are in 2.50 moles of of Helium gas at STP? 2.50 mol 22.4 L 1 mol = 56.0 L He C. Molar Conversion Examples Find the mass of 137 L of SO2 gas at STP. 137 L SO2 1 mol 64.07 g 22.4 L SO2 1 mol = 392 g SO2 Find the number of Liters of CO2 at STP if it has a mass of 132 grams.